# Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt = k*v^2.?

Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt = k*v^2. The initial speed of the motorboat is v(0) = 10 m/s, and v is decreasing at the rate of 1 m/s^2 when v=5 m/s. (a) How long does it take for the velocity of the boat to decrease to 1 m/s? (b) To (1/10) m/s? (c) When does the boat come to a stop?

Thanks

### 1 Answer

- hfshawLv 71 decade agoBest Answer
You have that:

dv/dt = k*v^2

and you should already be able to tell that k is going to be a negative number.

You are told that when v = 5 m/s, dv/dt = -1 m/s^2, so:

-1 m/s^2 = k*(5 m/s)^2

k = -1/(25 m)

The original differential equation is separable:

v^(-2) dv = k dt

which integrates to:

-1/v = k*t + c

where c is the constant of integration.

v(t) = -1/(k*t+c)

You are told that at t = 0, v(t) = 10m/s, so:

10 m/s = -1/c

c = -0.1 s/m

Plugging in the values for k and c, we have that the equation for the velocity is:

v(t) = 1/(0.1 s/m + t/25m)

The velocity of the boat is 1 m/s when:

1m/s = 1/(0.1 s/m + t/25m)

1s/m = (0.1 s/m + t/25m)

25 s = 2.5s + t

t = 22.5 sec

The velocity of the boat is 0.1 m/s when:

0.1 m/s = 1/(0.1 s/m + t/25m)

10 s/m = 0.1 s/m + t/25m

250s = 2.5 s + t

t = 247.5 sec

The boat's speed is zero only in the limit as t -> infinity.